of 17 /17
64 Speed Control of DC Motors: the speed of a motor is given by the relation a a a a R I V K p A Z R I V N 60 where R a = armature circuit resistance. It is obvious that the speed can be controlled by varying (i) Flux/pole, Φ (Flux Control) (ii) Resistance R a of armature circuit (Rheostatic Control) and (iii) Applied voltage V (Voltage Control). Speed Control of Shunt motor: (i) Variation of Flux or Flux Control Method: By decreasing the flux, the speed can be increased and vice versa. The flux of a dc motor can be changed by changing I sh with help of a shunt field rheostat. Since I sh is relatively small, shunt field rheostat has to carry only a small current, which means I 2 sh R loss is small, so that rheostat is small in size. (ii) Armature or Rheostatic Control Method: This method is used when speeds below the no-load speed are required. As the supply voltage is normally constant, the voltage across the armature is varied by inserting a variable rheostat in series with the armature circuit. As controller resistance is increased, voltage across the armature is decreased, thereby decreasing the armature speed. For a load constant torque, speed is approximately proportional to the voltage across the armature. From the

trinhnga
• Category

## Documents

• view

249

2

Embed Size (px)

### Transcript of Speed Control of DC Motors: the speed of a motor is given by the ... 64

Speed Control of DC Motors: the speed of a motor is given by the

relation

aaaa RIVK

pA

ZRIVN

60 where Ra= armature circuit resistance. It is

obvious that the speed can be controlled by varying

(i) Flux/pole, Φ (Flux Control)

(ii) Resistance Ra of armature circuit (Rheostatic Control) and

(iii) Applied voltage V (Voltage Control).

Speed Control of Shunt motor:

(i) Variation of Flux or Flux Control Method: By decreasing the flux, the

speed can be increased and vice versa. The flux of a dc motor can be

changed by changing Ish with help of a shunt field rheostat. Since Ish is

relatively small, shunt field rheostat has to carry only a small current,

which means I2shR loss is small, so that rheostat is small in size.

(ii) Armature or Rheostatic Control Method: This method is used when

speeds below the no-load speed are required. As the supply voltage is

normally constant, the voltage across the armature is varied by inserting a

variable rheostat in series with the armature circuit. As controller

resistance is increased, voltage across the armature is decreased, thereby

decreasing the armature speed. For a load constant torque, speed is

approximately proportional to the voltage across the armature. From the 65

speed/armature current characteristic, it is seen that greater the resistance

in the armature circuit, greater is the fall in the speed.

(iii) Voltage Control Method:

(a) Multiple Voltage Control: In this method, the shunt field of the motor

is connected permanently to a fixed exciting voltage, but the armature is

supplied with different voltages by connecting it across one of the several

different voltages by means of suitable switchgear. The armature speed

will be approximately proportional to these different voltages. The

intermediate speeds can be obtained by adjusting the shunt field regulator.

(b) Ward-Leonard System: This system is used where an unusually wide

and very sensitive speed control is required as for colliery winders,

electric excavators, elevators and the main drives in steel mills and

blooming and paper mills. M1 is the main motor whose speed control is

required. The field of this motor is permanently connected across the dc

supply lines. By applying a variable voltage across its armature, any

desired speed can be obtained. This variable voltage is supplied by a

motor-generator set which consists of either a dc or an ac motor M2

directly coupled to generator G. The motor M2 runs at an approximately

constant speed. The output voltage of G is directly fed to the main motor

M1. The voltage of the generator can be varied from zero up to its

maximum value by means of its field regulator. By reversing the direction 66

of the field current of G by means of the reversing switch RS, generated

voltage can be reversed and hence the direction of rotation of M1. It

should be remembered that motor generator set always runs in the same

direction.

Speed Control of Series Motors:

1. Flux Control Method: Variations in the flux of a series motor can be

brought about in any one of the following ways:

(a) Field Diverters: The series winding are shunted by a variable

resistance known as field diverter. Any desired amount of current can be

passed through the diverter by adjusting its resistance. Hence the flux can

be decreased and consequently, the speed of the motor increased.

(b) Armature Diverter: A diverter across the armature can be used for

giving speeds lower than the normal speed. For a given constant load

torque, if Ia is reduced due to armature diverter, the must increase

(∵Ta aI ) This results in an increase in current taken from the supply

(which increases the flux and a fall in speed (N I/ )). The variation in

speed can be controlled by varying the diverter resistance. 67

(c) Trapped Field Control Field: This method is often used in electric

traction. The number of series filed turns in the circuit can be changed.

With full field, the motor runs at its minimum speed which can be raised

in steps by cutting out some of the series turns.

(d) Paralleling Field coils: this method used for fan motors, several

speeds can be obtained by regrouping the field coils. It is seen that for a

4-pole motor, three speeds can be obtained easily. 68

2. Variable Resistance in Series with Motor: By increasing the resistance

in series with the armature the voltage applied across the armature

terminals can be decreased. With reduced voltage across the armature, the

speed is reduced. However, it will be noted that since full motor current

passes through this resistance, there is a considerable loss of power in it.

Problems 41- A 230 V dc shunt motor runs at 800 rpm and takes armature current of 50 A.

Find resistance to be added to the field circuit to increase speed to 1000 rpm

at an armature current of 80 A. Assume flux proportional to field current.

Armature resistance = 0.15 Ω and field winding resistance = 250 Ω.

2- A 250 V, dc shunt motor has an armature resistance of 0.5 Ω and a field

resistance of 250 Ω. When driving a load of constant torque at 600 rpm, the

armature current is 20 A. If it is desired to raise the speed from 600 to 800

rpm, what resistance should be inserted in the shunt field circuit? Assume that

the magnetic circuit is unsaturated.

3- A dc shunt motor takes an armature current of 20 A from a 220 V supply.

Armature circuit resistance is 0.5 ohm. For reducing the speed by 50%,

calculate the resistance required in the series, with the armature, if (a) the

load torque is constant (b) the load torque is proportional to the square of the

speed.

4- A 7.48 kW, 220 V, 990 rpm shunt motor has a full load efficiency of 88%, the

armature resistance is 0.08 ohm and shunt field current is 2 A. If the speed of

this motor is reduced to 450 rpm by inserting a resistance in the armature 69

circuit, find the motor output, the armature current, external resistance to be

inserted in the armature circuit and overall efficiency. Assume the load torque

to remain constant.

5- A 250 V dc shunt motor has armature circuit resistance of 0.5 Ω and a field

circuit resistance of 125 Ω. It drives a load at 1000 rpm and takes 30 A. The

field circuit resistance is then slowly increased to 150 Ω. If the flux and field

current can be assumed to be proportional and if the load torque remains

constant, calculate the final speed and armature current. [1186 rpm. 33.6 A]

6- A shunt-wound motor has a field resistance of 400 Ω and an armature

resistance of 0.1 Ω and runs off 240 V supply. The armature current is 60 A

and the motor speed is 900 rpm; Assuming a straight line magnetization

curve, calculate (a) the additional resistance in the field to increase the speed

to 1000 rpm for the same armature current and (b) the speed with the original

field current of 200 A. [(a) 44.4 Ω (b) 842.5 rpm]

7- A 250-V shunt motor has an armature current of 20 A when running at 1000

rpm against full load torque. The armature resistance is 0.5 Ω. What

resistance must be inserted in series with the armature to reduce the speed to

500 rpm at the same torque and what will be the speed if the load torque is

halved with this resistance in the circuit? Assume the flux to remain constant

throughout and neglect brush contact drop.

8- A 7.46 kW, 220 V, 900 rpm shunt motor has a full-load efficiency of 88 per

cent, an armature resistance of 0.08 Ω and shunt field current of 2 A. If the

speed of this motor is reduced to 450 rpm by inserting a resistance in the

armature circuit, the load torque remaining constant, find the motor output,

the armature current, the external resistance and the overall efficiency.

9- A 200 V, dc series motor takes 40 A when running at 700 rpm. Calculate the

speed at which the motor will run and the current taken from the supply if the

field is shunted by a resistance equal to the field resistance and the load

torque is increased by 50%. Armature resistance = 0.15 Ω, field resistance =

0.1 Ω It may be assumed that flux per pole is proportional to the field.

10- A 4-pole, 250 V dc series motor takes 20 A and runs at 900 rpm each field

coil has resistance of 0.025 ohm and the resistance of the armature is 0.1 ohm.

At what speed will the motor run developing the same torque if : (i) a diverter

of 0.2 ohm is connected in parallel with the series field (ii) rearranging the 70

field coils in two series and parallel groups Assume unsaturated magnetic

operation.

11- A 200V, dc series motor runs at 500rpm when taking a line current of 25A.

The resistance of the armature is 0.2Ω and that of the series field 0.6Ω. At

what speed will it run when developing the same torque when armature

diverter of 10Ω is used? Assume a straight line magnetization curve.[314rpm]

Electric Braking: A motor and its load may be brought to rest quickly by

using either (i) Friction Braking or (ii) Electric Braking. Mechanical

brake has one drawback: it is difficult to achieve a smooth stop because

it depends on the condition of the braking surface as well as on the skill

of the operator. The excellent electric braking methods are available

which eliminate the need of brake lining levers and other mechanical

gadgets. Electric braking, both for shunt and series motors, is of the

following three types:

(i) Rheostatic or dynamic braking

(ii) Plugging i.e., reversal of torque so that armature tends to rotate

in the opposite direction.

(iii) Regenerative braking.

Obviously, friction brake is necessary for holding the motor even after it

has been brought to rest.

Electric Braking of Shunt Motors:

(a) Rheostatic or Dynamic Braking: In this method, the armature of

the shunt motor is disconnected from the supply and is connected

across a variable resistance R. The field winding is left connected

across the supply. The braking effect is controlled by varying the

series resistance R. Obviously, this method makes use of generator

action in a motor to bring it to rest. 71

(b) Plugging or Reverse Current Braking: This method is commonly

used in controlling elevators, rolling mills, printing presses and

machine tools etc. In this method, connections to the armature

terminals are reversed so that motor tends to run in the opposite

direction. Due to the reversal of armature connections, applied

voltage V and E start acting in the same direction around the

circuit. In order to limit the armature current to a reasonable

value, it is necessary to insert a resistor in the circuit while

reversing armature connections.

(c) Regenerative Braking: This method is used when the load on the

motor has over-hauling characteristic as in the lowering of the

cage of a hoist or the downgrade motion of an electric train.

Regeneration takes place when Eb becomes grater than V. This

happens when the overhauling load acts as a prime mover and so

drives the machines as a generator. Consequently, direction of Ia

and hence of armature torque is reversed and speed falls until E

becomes lower than V. It is obvious that during the slowing down

of the motor, power is returned to the line which may be used for 72

supplying another train on an upgrade, thereby relieving the

powerhouse of part of its load.

Electric Braking of Series Motor:

(a) Rheostatic (or dynamic) Braking: The motor is disconnected from

the supply, the field connections are reversed and the motor is

connected in series with a variable resistance R. Obviously, now,

the machine is running as a generator. The field connections are

reversed to make sure that current through field winding flows in

the same direction as before (i.e., from M to N ) in order to assist

residual magnetism. In practice, the variable resistance employed

for starting purpose is itself used for braking purposes.

(b) Plugging or Reverse Current Braking: As in the case of shunt

motors, in this case also the connections of the armature are

reversed and a variable resistance R is put in series with the

armature.

(c) Regenerative Braking: This type of braking of a series motor is not

possible without modification because reversal of Ia would also

mean reversal of the field and hence of Eb. However, this method 73

is sometimes used with traction motors, special arrangements

being necessary for the purpose.

Starter of DC Motors: the current drawn by a motor armature is given

by the relation where V is the supply voltage, the back emf and R the

armature resistance. When the motor is at rest, there is no back emf

developed in the armature. If, now, full supply voltage is applied across

the stationary armature, it will draw a very large current because armature

resistance is relatively small. This excessive current will blow out the

fuses and, prior to that, it will damage the commutator and brushes etc.

To avoid this happening, a resistance is introduced in series with the

armature (for the duration of starting period only, say 5 to 10 seconds)

which limits the starting current to a safe value. The starting resistance is

gradually cut out as the motor gains speed and develops the back emf

which then regulates its speed. Very small motors may, however, be

started from rest by connecting them directly to the supply lines. It does

not result in any harm to the motor for the following reasons:

1- Such motors have a relatively higher armature resistance than large

motors; hence their starting current is not so high.

2- Being small, they have low moment of inertia, hence they speed up

quickly.

3- The momentary large starting current taken by them is not sufficient to

produce a large disturbance in the voltage regulation of the supply lines.12- A 10 hp (7.46kW) 200 V shunt motor has full-load efficiency of 85%. The

armature has a resistance of 0.25Ω. Calculate the value of the starting

resistance necessary to limit the starting current to 1.5 times the full-load

current at the moment of first switching on. The shunt current may be

neglected. Find also the back emf of the motor, when the current has fallen to

its full load value, assuming that the whole of the starting resistance is still in

circuit. 74

Swinburne's Test: It is a simple method in which losses are measured

separately and from their knowledge, efficiency at any desired load can

test. However, this test is applicable to those machines in which flux is

practically constant i.e. shunt and compound-wound machines. The

machine is running as a motor on no-load at its rated voltage. The speed

is adjusted to the rated speed with the help of shunt regulator. The no-

load current I is measured by the ammeter A1 whereas shunt field current

Ish is given by ammeter A2. The no-load armature current is (I0-Ish) or Ia0.

Let, supply voltage = V, No-load input power = VI0

Power input to armature=V(I0−Ish); Power input to shunt = VIsh

No-load power input to armature supplies the following losses : (i) Iron

losses in core (ii) friction loss (iii) windage loss and(iv) armature Cu

loss,(I0−Ish)2Ra or Ia02Ra

In calculating armature Cu loss, ‘hot’ resistance of armature should be

used. A stationary measurement of armature circuit resistance at the

room-temperature (15Co) is made by passing current through the armature

from a low voltage dc supply.

Then, the ‘hot’ resistance, allowing a temperature rise of 50Co is found

thus: R15 = Ro (1 + 15α0) ; R65 =Ro (1 + 65α0),o

oRR

151651

1565

If we subtract from the total input the no-load armature Cu loss, then we

get constant losses. 75

Constant losses Wc= VI0−(I0−Ish)2Ra

Knowing the constant losses of the machine, its efficiency at any other

load can be determined as given below. Let I = load current at which

efficiency is required.

Then, armature current is Ia= I − Ish ...if machine is motoring

= I + Ish ...if machine is generating

Efficiency when running as a motor

Input = VI, Armature Cu loss = Ia2Ra=(I-Ish)2Ra

Constant losses = Wc

Total losses = (I−Ish)2Ra+Wc; VIWRIIVI

inputlossesinput cash

m

2)(

Efficiency when running as a generator

Output = VI ; Armature Cu loss = (I + Ish)2Ra; constant loss=Wc

Total losses = (I+Ish)2R+Wc;cash

g WRIIVIVI

lossesoutputoutput

2)(

1. It is convenient and economical because power required to test a large

machine is small i.e. only no-load input power.

2. The efficiency can be predetermined at any load because constant-

losses are known.

1. No account is taken of the change in iron losses from no-load to full-

increases the iron losses in some cases by as much as 50%.

2. As the test is on no-load, it is impossible to know whether

commutation would be satisfactory at full-load and whether the

temperature rise would be within the specified limits. 76

Regenerative or Hopkinson's Test (Back to Back Test): By this

method, full-load test can be carried out on two shunt machines,

preferably identical ones, without wasting their outputs. The two

machines are mechanically coupled and are so adjusted electrically that

one of them runs as a motor and the other as a generator. The mechanical

output of the motor drives the generator and the electrical output of

generator is used in supplying the greater part of input to the motor. If

there were no losses in the machines, they would have run without any

external power supply. But due to these losses, generator output is not

sufficient to drive the motor and vice-versa. The losses are supplied either

by an extra motor which is belt-connected to the motor-generator set, or

by electrically from the supply mains. The two shunt machines are

connected in parallel. They are, to begin with, started as unloaded motors.

Then, the field of one is weakened and that of the other is strengthened so

that the former runs as a motor and the latter as a generator. The usual

method of procedure is as follows: Machine M is started up from the

supply mains with the help of a starter (not shown) whereas main switch

S of the other machine is kept open. Its speed is adjusted to normal value

by means of its field regulator. Machine M drives machine G as a

generator and its voltage is read on voltmeter V. The voltage of G is

showing that its voltage is the same, both in polarity and magnitude as

that of the main supply. Thereafter, S is closed to parallel the machines.

By adjusting the respective field regulators, any load can now be thrown

on to the machines. Generator current I1 can be adjusted to any desired

value by increasing the excitation of G or by reducing the excitation of M

and the corresponding values of different ammeters are read. The

electrical output of the generator plus the small power taken from the 77

supply, is taken by the motor and is given out as a mechanical power after

supplying the motor losses. If supply voltage is V, then

Motor input=V(I1+I2), where I2 is the current taken from the supply.

Generator output = VI1

Assuming that both machines have the same efficiency η.

Output of motor = η × input = η V(I1+I2) = generator input

Output of generator=η×input=η×ηV(I1+I2)=η2V(I1 + I2)

η2V(I1 + I2)=VI1 or η =21

1

III

However, it is not quite correct to assume equal efficiencies for two

machines because their armature currents as well as excitations are

different. We will not find the efficiencies separately.

Let Ra= armature resistance of each machine

I3= exciting current of the generator

I4= exciting current of the motor

Armature Cu loss in generator = (I1+I3)2Ra

Armature Cu loss in motor = (I1+I2-I4)2Ra

Shunt Cu loss in generator = VI3

Shunt Cu loss in motor = VI4 78

But total motor and generator losses are equal to the power supplied by

the mains. Power drawn from supply = VI2

If we subtract the armature and shunt Cu losses from this, we get the stray

losses of both machines.

Total stray losses for the set= VI2−[(I1+I3)2Ra+(I1+I2−I4)2Ra+VI3+VI4]=W

Making one assumption that stray losses are equally divided between the

two machines, we have Stray loss per machine = W/2

For Generator

Total losses = (I1+I3)2Ra+VI3+W/2=Wg

Output=VI1g

g WVIVI

1

1

For Motor

Total losses = (I1+I2-I4)2Ra+VI4+W/2=Wm

Input = V(I1+I2) )()(

21

21

IIVWIIV m

m

Alternative Connections for Hopkinson’s Test: the following figure

shows in slightly different method of connecting the two machines to the

supply. Here, the main difference is that the shunt windings are directly

connected across the lines. Hence, the line input current is I1 excluding

the field currents. The efficiencies can be calculated as detailed below:

Motor armature Cu loss = (I1+I2)2Ra

Generator armature Cu loss = I22Ra

Power drawn from the supply =Winput =VI1

Total stray losses i.e. iron, friction and windage losses for the two

machines are = VI1−[(I1+I2)2Ra− I22 Ra] = W

Stray loss for each machine = W/2

Motor Efficiency

Motor input = armature input + shunt field input = V ( I1+I2)+VI3+Winput 79

Motor losses=armature Cu loss+shunt Cu loss+stray losses

=(I1+I2)2Ra+VI3+W/2=Wm

Motor efficiency = η = %100

input

minput

WWW

Generator Efficiency:

Generator output = VI2

Generator losses = I22RaVI4+W/2=Wg

Generator efficiency=gWVI

VI

2

2

1. Power required for the test is small as compared to the full-load

powers of the two machines.

2. As machines are being tested under full-load conditions, the

temperature rise and the commutation qualities of the machines can

be observed.

3. Because of full-load conditions, any change in iron loss due to flux

distortion at full-load, is being taken into account.

The only disadvantage is with regard to the availability of two identical

machines. 80

13- The no-load test of a 44.76 kW, 220 V dc shunt motor gave thefollowing figures: Input current=13.25 A; field current=2.55 A;resistance of armature at 75Co = 0.032Ω and brush drop =2 V.Estimate the full-load current and efficiency.[90%]

14- A 250V, 14.92 kW shunt motor has a maximum efficiency of 88% anda speed of 700 rpm when delivering 80% of its rated output. Theresistance of its shunt field is 100Ω. Determine the efficiency andspeed when the motor draws a current of 78 A from the mains.[ 86.9%,680 rpm]

15- The Hopkinson’s test on two similar shunt machines gave thefollowing full-load data: Line voltage=110V Field currents are 3A and3.5A Line current = 48A Armature resistance of each is 0.035Ω Motorarmature current=230A Calculate the efficiency of each machineassuming a brush contact drop of 1 volt per brush.[ 88.8%, 89.4%]

16- The Hopkinson’s test on two shunt machines gave the following resultsfor full-load: Line voltage=250V; current taken from supply systemexcluding field currents =50A; motor armature current = 380 A; fieldcurrents 5A and 4.2A. Calculate the efficiency of the machine workingas a generator. Armature resistance of each machine is 0.2 Ω.[92.02%]